Question

The driving force for fluid flow is the pressure difference, and a pump operates by raising the pressure of a fluid (by converting the mechanical shaft work to flow energy). A gasoline pump is measured to consume 3.8 kW of electric power when operating, If the pressure differential between-the outlet and inlet of the pump is measured to be 7 kPa and the changes in velocity and elevation are negligible, determine the maximum possible volume flow rate of gasoline.

Answer 1

Answer:

$\dot V = 0.542 \frac{m^{3}}{s}$

Explanation:

The power needed for the pump to raise the pressure of gasoline is defined by following equation. The maximum possible volume flow rate is isolated and then calculated:

$\dot W = \dot V \cdot \Delta P\\\dot V = \frac{\dot W}{\Delta P}\\\dot V = \frac{3.8 kW}{7 kPa}\\\dot V = 0.542 \frac{m^{3}}{s}$

Answer 2

Explanation:

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